$\int \sqrt[5]{x^2}\,dx=$ $+C$
At first it might seem as if we can't apply any rule we've learned to find the indefinite integral of a radical function. However, remember that any radical can be rewritten as a rational power. $\int \sqrt[5]{x^2}\,dx=\int x^{^{\frac25}}\,dx$ Now we can integrate using the reverse power rule: $\int x^n\,dx=\dfrac{x^{n+1}}{n+1}+C$ $\begin{aligned} \int \sqrt[5]{x^2}\,dx&=\int x^{^{\frac{2}{5}}}\,dx \\\\ &=\dfrac{x^{^{\frac{2}{5}+1}}}{\dfrac{2}{5}+1}+C \\\\ &=\dfrac{5}{7} x^{^{\frac{7}{5}}}+C \end{aligned}$ In conclusion, $\int \sqrt[5]{x^2}\,dx=\dfrac{5}{7} x^{^{\frac{7}{5}}}+C$